This is an example of a word problem in algebra which can be solved by expressing the information given into a one variable equation. 2/3 of 300 is equal to 200, so the obvious answer is that Paula after spending her money had $200. The equation to solve this problem could be expressed as X - 1/9 = 200. And after solving this equation you would find that the original amount of money Paula had was $211.
The key to solving algebraic word problems is to read the question carefully and determine just what it is that is being asked for. Most of these problems are seeking the answer to a single variable, in the case of this problem, just how much money did Paula start out with? You get a clue with the answer of how much money she had left after her shopping spree, which in this case is $200, which is derived by taking Bob's amount of money and dividing it into thirds, and then subtracting 1/3 of the amount.
Once you have all of the numbers except one, you can express them in the form of an algebra equation and solve for the one remaining variable. These types of problems cause more people to fail algebra than any other part of the course. The key to solving them is to remain calm and to make sure you take in all of the information that you are given, and solve the easy parts before you get to the part of devising the equation.
Take your time, and read through the entire problem completely before you start to work it. Remember that to get a value from one side to the other that is negative, you will make it positive on the other side. If a number on one side is bring divided, to move it to the other side will make it a multiplicand.
The key to solving algebraic word problems is to read the question carefully and determine just what it is that is being asked for. Most of these problems are seeking the answer to a single variable, in the case of this problem, just how much money did Paula start out with? You get a clue with the answer of how much money she had left after her shopping spree, which in this case is $200, which is derived by taking Bob's amount of money and dividing it into thirds, and then subtracting 1/3 of the amount.
Once you have all of the numbers except one, you can express them in the form of an algebra equation and solve for the one remaining variable. These types of problems cause more people to fail algebra than any other part of the course. The key to solving them is to remain calm and to make sure you take in all of the information that you are given, and solve the easy parts before you get to the part of devising the equation.
Take your time, and read through the entire problem completely before you start to work it. Remember that to get a value from one side to the other that is negative, you will make it positive on the other side. If a number on one side is bring divided, to move it to the other side will make it a multiplicand.
